7.3: Exercises

Exercise \(\PageIndex{1}\)

Use the horizontal line test to determine if the function is one-to-one.

1. 
2. 
3. \(f(x)=x^2+2x+5\)
4. \(f(x)=x^2-14 x+29\)
5. \(f(x)=x^3-5x^2\)
6. \(f(x)=\dfrac{x^2}{x^2-3}\)
7. \(f(x)=\sqrt{x+2}\)
8. \(f(x)=\sqrt{|x+2|}\)

Answer

1. no (that is: the function is not one-to-one)
2. yes
3. no
4. no
5. no
6. no
7. yes
8. no

Exercise \(\PageIndex{2}\)

Find the inverse of the function \(f\) and check your solution.

1. \(f(x)=4x+9\)
2. \(f(x)=-8x-3\)
3. \(f(x)=\sqrt{x+8}\)
4. \(f(x)=\sqrt{3x+7}\)
5. \(f(x)=6\cdot \sqrt{-x-2}\)
6. \(f(x)=x^3\)
7. \(f(x)=(2x+5)^3\)
8. \(f(x)=2\cdot x^3+5\)
9. \(f(x)=\dfrac{1}{x}\)
10. \(f(x)=\dfrac{1}{x-1}\)
11. \(f(x)=\dfrac{1}{\sqrt{x-2}}\)
12. \(f(x)=\dfrac{-5}{4-x}\)
13. \(f(x)=\dfrac{x}{x+2}\)
14. \(f(x)=\dfrac{3x}{x-6}\)
15. \(f(x)=\dfrac{x+2}{x+3}\)
16. \(f(x)=\dfrac{7-x}{x-5}\)
17. \(f\) given by the table: \(\begin{array}{|c||c|c|c|c|c|c|}
    \hline x & 2 & 4 & 6 & 8 & 10 & 12 \\
    \hline \hline f(x) & 3 & 7 & 1 & 8 & 5 & 2 \\
    \hline
    \end{array} \)

Answer

1. \(f^{-1}(x)=\dfrac{x-9}{4}\)
2. \(f^{-1}(x)=-\dfrac{x+3}{8}\)
3. \(f^{-1}(x)=x^{2}-8\)
4. \(f^{-1}(x)=\dfrac{x^{2}-7}{3}\)
5. \(f^{-1}(x)=-\left(\dfrac{x}{6}\right)^{2}-2=\dfrac{-x^{2}-72}{36}\)
6. \(f^{-1}(x)=\sqrt[3]{x}\)
7. \(f^{-1}(x)=\dfrac{\sqrt[3]{x}-5}{2}\)
8. \(f^{-1}(x)=\sqrt[3]{\dfrac{x-5}{2}}\)
9. \(f^{-1}(x)=\dfrac{1}{x}+1=\dfrac{1+x}{x}\)
10. \(f^{-1}(x)=\left(\dfrac{1}{x}\right)^{2}+2=\dfrac{1+2 x^{2}}{x^{2}}\)
11. \(f^{-1}(x)=\dfrac{5}{y}+4=\dfrac{5+4 y}{y}\)
12. \(f^{-1}(x)=\dfrac{5}{y}+4=\dfrac{5+4 y}{y}\)
13. \(f^{-1}(x)=\dfrac{2 x}{1-x}\)
14. \(f^{-1}(x)=\dfrac{6 x}{x-3}\)
15. \(f^{-1}(x)=\dfrac{2-3 x}{x-1}\)
16. \(f^{-1}(x)=\dfrac{5 x+7}{x+1}\)
17. \(\begin{array}{|c||c|c|c|c|c|c|}
    \hline x & 3 & 7 & 1 & 8 & 5 & 2 \\
    \hline \hline f^{-1}(x) & 2 & 4 & 6 & 8 & 10 & 12 \\
    \hline
    \end{array}\)

Exercise \(\PageIndex{3}\)

Restrict the domain of the function \(f\) in such a way that \(f\) becomes a one-to-one function. Find the inverse of \(f\) with the restricted domain.

1. \(f(x)=x^2\)
2. \(f(x)=(x+5)^2+1\)
3. \(f(x)=|x|\)
4. \(f(x)=|x-4|-2\)
5. \(f(x)=\dfrac{1}{x^2}\)
6. \(f(x)=\dfrac{-3}{(x+7)^2}\)
7. \(f(x)=x^4\)
8. \(f(x)=\dfrac{(x-3)^4}{10}\)

Answer

1. restricting to the domain \(D=[0, \infty)\) gives the inverse \(f^{-1}(x)=\sqrt{x}\)
2. restricting to the domain \(D=[-5, \infty)\) gives the inverse \(f^{-1}(x)=\sqrt{x-1}-5\)
3. restricting to the domain \(D=[0, \infty)\) gives the inverse \(f^{-1}(x)=x\)
4. restricting to the domain \(D=[4, \infty)\) gives the inverse \(f^{-1}(x)=x+6\)
5. restricting to the domain \(D=[0, \infty)\) gives the inverse \(f^{-1}(x)=\sqrt{\dfrac{1}{x}}\)
6. restricting to the domain \(D=[-7, \infty)\) gives the inverse \(f^{-1}(x)=\sqrt{-\dfrac{3}{x}}-7\)
7. restricting to the domain \(D=[0, \infty)\) gives the inverse \(f^{-1}(x)=\sqrt[4]{x}\)
8. restricting to the domain \(D=[3, \infty)\) gives the inverse \(f^{-1}(x)=3+\sqrt[4]{10 x}\)

Exercise \(\PageIndex{4}\)

Determine whether the following functions \(f\) and \(g\) are inverse to each other.

1. \(f(x)=x+3, \quad g(x)=x-3\),
2. \(f(x)=-x-4, \quad g(x)=4-x\),
3. \(f(x)=2x+3, \quad g(x)=x-\dfrac{3}{2}\),
4. \(f(x)=6x-1, \quad g(x)=\dfrac{x+1}{6}\),
5. \(f(x)=x^3-5, \quad g(x)= 5+\sqrt[3]{x}\),
6. \(f(x)=\dfrac{1}{x-2}, \quad g(x)=\dfrac{1}{x}+2\).

Answer

1. yes (that is: the functions f and g are inverses of each other)
2. no
3. no
4. yes
5. no
6. yes

Exercise \(\PageIndex{5}\)

Draw the graph of the inverse of the function given below.

1. 
2. 
3. 
4. \(f(x)=\sqrt{x}\)
5. \(f(x)=x^3-4\)
6. \(f(x)=2x-4\)
7. \(f(x)=2^x\)
8. \(f(x)=\dfrac{1}{x-2}\) for \(x>2\)
9. \(f(x)=\dfrac{1}{x-2}\) for \(x<2\)

Answer

1. 
2. 
3. 
4. 
5. 
6. 
7. 
8. 
9.